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Expert-verified Found in: Page 108 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # A warehouse is being built that will have neither heating nor cooling. Depending on the amount of insulation, the time constant for the building may range from 1 to 5 hr. To illustrate the effect insulation will have on the temperature inside the warehouse, assume the outside temperature varies as a sine wave, with a minimum of ${\mathbf{16}}{\mathbf{°}}{\mathbf{C}}$ at ${\mathbf{2}}{\mathbf{:}}{\mathbf{00}}{\mathbf{a}}{\mathbf{.}}{\mathbf{m}}{\mathbf{.}}$ and a maximum of ${\mathbf{32}}{\mathbf{°}}{\mathbf{C}}$ at ${\mathbf{2}}{\mathbf{:}}{\mathbf{00}}{\mathbf{p}}{\mathbf{.}}{\mathbf{m}}{\mathbf{.}}$ Assuming the exponential term (which involves the initial temperature T0) has died off, what is the lowest temperature inside the building if the time constant is 1 hr? If it is 5 hr? What is the highest temperature inside the building if the time constant is 1 hr? If it is 5 hr?

If the time constant is 1 hour, the lowest temperature inside the building will reach $16.3°\mathrm{C}$and the highest temperature will reach $31.7°\mathrm{C}$.

If the time constant is 5 hours, the lowest temperature inside the building will reach $19.1°\mathrm{C}$ and the highest temperature will reach $28.{9}^{\mathrm{o}}\mathrm{C}$.

See the step by step solution

## Step1: Given data.

The temperature outside the building varies as a sine wave, with a minimum of $16°\mathrm{C}$ at $2:00\mathrm{a}.\mathrm{m}.$and a maximum of $32{}^{\circ }\mathrm{C}$at $2:00\mathrm{p}.\mathrm{m}.$ If the time constants for the building are 1 hours and 5 hours, it has to find the lowest and highest temperatures inside the building.

## Step 2: Analyzing the given statement

$M={M}_{0}-Bcos\left(\omega t\right)$ …… (1)

${M}_{0}-B=16 ......\left(a\right)\phantom{\rule{0ex}{0ex}}{M}_{0}+B=32 ......\left(b\right)$

At $2:00\mathrm{a}.\mathrm{m}.$t=0,

And at 2:00 p.m, t=12 hours

Adding the equations (a) and (b),

$2{M}_{0}=48\phantom{\rule{0ex}{0ex}}{M}_{0}=24\phantom{\rule{0ex}{0ex}}$

Therefore, B=8.

## Step3: To determine the temperature at time t.

The forcing function $\mathrm{Q}\left(\mathrm{t}\right)$is given by,

$Q\left(t\right)=K\left({M}_{0}-Bcos\left(\omega t\right)\right)\phantom{\rule{0ex}{0ex}}Q\left(t\right)=K\left(24-8cos\left(\omega t\right)\right)\phantom{\rule{0ex}{0ex}}$

Temperature T(t) is given by

$T\left(t\right)={B}_{0}-BF\left(t\right)+C{e}^{-kt}$ …… (2)

Where, $F\left(t\right)=\frac{cos\left(\omega t\right)+\left(\frac{\omega }{k}\right)sin\left(\omega t\right)}{1+\frac{{\omega }^{2}}{{k}^{2}}}$.

Substituting K=1,B=8 and B0 =M0 =24 in equation (2),

$T\left(t\right)=24-8F\left(t\right)+C{e}^{-t}$

Now as the exponential term died off, therefore,

$T\left(t\right)=24-8F\left(t\right)$ …… (3)

Where, the value of F(t) is,

$F\left(t\right)=\frac{1}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}\left(\frac{cos\left(\omega t\right)}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}+\frac{\left(\frac{\omega }{k}\right)sin\left(\omega t\right)}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}\right)$

$F\left(t\right)=\frac{sin\left(\omega t+ta{n}^{-1}\left(\frac{k}{\omega }\right)\right)}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}$ …… (4)

## Step 4: To determine lowest and highest temperatures inside the building if the time constant is 1 hr

Now as the maximum value of sin x is 1.

Therefore, from equation (4),

$F\left(t\right)=\frac{1}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}$

So, by substituting the value of F(t) in equation (3),

$T\left(t\right)=24-\frac{8}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}$

…… (5)

When the time constant is 1 hour i.e., when $\frac{1}{\mathrm{k}}=1$

And $\omega =\frac{\pi }{12}$

Therefore, from equation (5),

Let ${T}_{L}$be the lowest temperature,

${T}_{L}=24-\frac{8}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}\phantom{\rule{0ex}{0ex}}{T}_{L}=24-\frac{8}{\sqrt{1+\frac{{\pi }^{2}}{144}}}\phantom{\rule{0ex}{0ex}}{T}_{L}=16.{3}^{0}C\phantom{\rule{0ex}{0ex}}$

Now as the minimum value of sin x is 1.

Therefore, from equation (4),

$F\left(t\right)=-\frac{1}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}$

Thus, from equation (5),

Let ${T}_{H}$be the highest temperature,

${T}_{H}=24+\frac{8}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}\phantom{\rule{0ex}{0ex}}{T}_{H}=24+\frac{8}{\sqrt{1+\frac{{\pi }^{2}}{144}}}\phantom{\rule{0ex}{0ex}}{T}_{H}=31.{7}^{0}C\phantom{\rule{0ex}{0ex}}$

Hence, if the time constant is 1 hour, the lowest temperature inside the building will reach role="math" localid="1664185402769" ${\mathbf{16}}{\mathbf{.}}{\mathbf{3}}{\mathbf{°}}{\mathbf{C}}$ and the highest temperature will reach role="math" localid="1664185416044" ${\mathbf{31}}{\mathbf{.}}{\mathbf{7}}{\mathbf{°}}{\mathbf{C}}$.

## Step 5: To determine lowest and highest temperatures inside the building if the time constant is 5 hr

When the time constant is 5 hours i.e., when$\frac{1}{\mathrm{K}}=\frac{1}{5}$

Hence, from equation (5),

Let ${T}_{L}$be the lowest temperature,

${T}_{L}=24-\frac{8}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}\phantom{\rule{0ex}{0ex}}{T}_{L}=24-\frac{8}{\sqrt{1+\frac{25{\pi }^{2}}{144}}}\phantom{\rule{0ex}{0ex}}{T}_{L}=19.{1}^{0}C\phantom{\rule{0ex}{0ex}}$

Let ${T}_{H}$be the highest temperature,

So, from equation (5),

${T}_{H}=24+\frac{8}{\sqrt{1+\frac{{\omega }^{2}}{{k}^{2}}}}\phantom{\rule{0ex}{0ex}}{T}_{H}=24+\frac{8}{\sqrt{1+\frac{25{\pi }^{2}}{144}}}\phantom{\rule{0ex}{0ex}}{T}_{H}=28.{9}^{0}C\phantom{\rule{0ex}{0ex}}$

Thereafter, if the time constant is 5 hours, the lowest temperature inside the building will reach $19.1°\mathrm{C}$ and the highest temperature will reach $28.9°\mathrm{C}$. ### Want to see more solutions like these? 